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Sum of binomial coefficients sum of powers of primes

The sum of all positive integers is
$$\sum_{n=1}^{\infty}n =1. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$
At least this is true for $n\leq 2$.
Can we generalize this to
$$\sum_{n=1}^{\infty}n^k =k \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad $$

A:

The answer is that for all $k$, we have the following identity:
$$\sum_{n=1}^{\infty}n^{k}=\frac{k(k-1)}{2} $$
Proof:
Consider a generating function of the form $$f_{k}(x)=\sum_{n=1}^{\infty}(x)_{n}^{k}$$
where
$(x)_{n}$ represents the falling factorial (i.e., $(x)_{n}=(x)(x-1)(x-2)…(x-n+1)$)
Now we can consider the following partial sum of $f
f30f4ceada

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